Learn

Friday, 9 October

JEE Main 2024 · PhysicsNumerical answerNumerical valueEasyCalculation

JEE Main 29 January 2024, Shift 1, Physics Q57

Question 57 of 90 in this shift, Physics question 27 of 30, Section B.

The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of 20 cm from its center is 1.5×10−5 T m1.5\times10^{-5}\ T\,m. The magnetic moment of the dipole is ______ A m2A\,m^2. (Given : μo4π=10−7 T m A−1\frac{\mu_o}{4\pi}=10^{-7}\ T\,m\,A^{-1})

Official answer

6

NTA final key.

Same idea in other shifts

Asked 2× in all
  1. 4 Apr 2026, Shift 2 · Q48Two identical small bar magnets each of dipole moment 353\sqrt5 J/T are placed at a center to center separation of 10 cm, with their axes…MediumNumerical valueHas a figure

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.